Is it possible to find the sum of all integer values that $x$ can take?

Question

Is it possible to find the sum of all integer values that $x$ can take? In: $$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$$

Answer 1

hint: $x+3-4\sqrt{x-1}=(\sqrt{x-1}-2)^2, x+8-6\sqrt{x-1}=(\sqrt{x-1}-3)^2$

Answer 2

Make the substitution $x=t^2+1$ $$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$$ $$\Rightarrow\sqrt{(t^2+1)+3-4\sqrt{(t^2+1)-1}}+\sqrt{(t^2+1)+8-6\sqrt{(t^2+1)-1}}=1$$ $$\Rightarrow\sqrt{t^2+1+3-4|t|}+\sqrt{t^2+1+8-6|t|}=1$$ $$\Rightarrow\sqrt{|t|^2-4|t|+4}+\sqrt{|t|^2-6|t|+9}=1$$ Make the substituion $|t|=a$ $$\Rightarrow\sqrt{(a-2)^2}+\sqrt{(a-3)^2}=1$$ $$\Rightarrow |a-2|+|a-3|=1$$ $$\Rightarrow |3-a|+|a-2|=1$$ We can see $(3-a)+(a-2)=1$ And we know that $$|a|+|b|\geq|a+b|$$ And equality holds when $a$ and $b$ have like sign. So we have, $$(3-a)(a-2)\geq 0$$ $$\Rightarrow (a-3)(a-2)\leq 0$$ $$\Rightarrow a\in [2,3] $$ Reverting back to $t$ $$\Rightarrow t\in [2,3]\cup [-3,-2] $$ Since $t=\sqrt{a-1}$, we have $$x\in [5,10] $$